M3 MS Jun 2006

June 2006 6679 Mechanics M3 Mark Scheme Question Number

Scheme

1.

Use of    y 2 dx  x     xy 2 dx

Marks

 x dx  x   x

2

M1

dx

1 2  1 3   2 x   x   3 x  16 64 x  2 3 8 x 3

Using limits 0 and 4

A1 = A1 M1 A1

(5) [5]

2.

(a)

Small Hemisphere Mass ratios

2 a   3 2

x

3 a 16

3

Bowl

2 7a  3 8

Large Hemisphere 3

2 3 a 3 Anything in the ratio 1 : 7 : 8 3 a 8

x

3 3 a  7  x  8 a 16 8 45 Leading to x a  112 1

(b)

Bowl M

Mass Ratios

45 a 112

x

M

Liquid kM

Bowl and Liquid  k  1 M

3 a 16

17 a 48

45 3 17 a  kM  a   k  1 M  a 112 16 48 2 Leading to k  7

B1

B1

M1 A1 cso

A1

(5)

B1 B1

M1 A1 A1

(5) [10]

1

Question Number

3.

Scheme

Marks

a  0.1 2 1     10  5 Fmax  ma 2

(a)

B1 M1 A1 M1

 0.2  0.1 10 

2

 N

 19.7

A1 (6)

cao a  0.2 ,

(b)

  10

v2   2  a 2  x2   100 2  0.22  0.12 

v  5.44

  3  ms 

2

B1ft, B1ft

 29.6 ...

1

cao If answers are given to more than 3 significant figures a maximum of one A mark is lost in the question.

4.

M1

tan  

3 4

tan  

r h

M1 A1 A1 (5) [11]

B1

or equivalent R r 3a or  h 4a

R 

r 5    R  mg  3   h



 mr 

B1

R sin   mg

M1 A1

R  

mg

R cos   mr 2

8g  10mrg  R   9a  9a 

tan  

M1 A1 A1

9a  5 10mrg   mg   8r  3 9a 

M1 A1

Eliminating R 3   3 9a  r  a   2   4 8r r 3a 4 h    2a tan  2 3

2

M1 A1 (11)

[11]

3

Question Number

5.

Scheme

(a)

Marks

A 0.75 m

B 1m

P AP    0.752  12  = 1.25 Conservation of energy 1 49  0.52  2  v2  2   2 g 1 1 2 2  0.75 for each incorrect term Leading to v  1.8  ms 1 

M1 A1

M1 A2 (1, 0)

A1

(6)

accept 1.81 (b) A 0.75 m

y T

B

 

T

P 2g R 

2T cos   2 g

0.75 sin  49  0.75  T  0.75   0.75  sin    1   49   1  sin   9.8  1   49   1 cos   sin  

M1 A1

y

Hooke’s Law

Eliminating T

tan   5 1  sin   5  tan   5sin  

cso

M1 A1

M1

A1

(6) [12]

4

5

Question Number

6.

Scheme

Marks

(a) v B1

Parabola 15

B1

Hyperbola

B1 (3)

Points 7.5

O

4

5

10 t

(b) Identifying the minimum point of the parabola and 5 as the end points.

A1 (2)

2t 5

(c) Splitting the integral into two part, with limits 0 and 4, and 4 and 5, and evaluating both integrals.



4 0

3t  t  4  dt  t 3  6t 2   32 and 4

0



5 4

3t  t  4  dt  t 3  6t 2   7

Both Total distance  39

5

4

 m

cso



(d)

t1 5

75 dt  32  7 t

75  ln t  51  25 t

ln

1 t1 1   t1  5e 3 5 3  6.98

cao

M1



M1 A1 A1 (3)

M1 A1 A1 M1 A1 (5) [13]

6

7

Question Number

7.

Scheme

Marks

(a) A

u

P

  5 gl 2

Conservation of Energy 1  5 gl  m  u 2   mgl 2  2   gl  Leading to u     2

M1 A1= A1



(b)

A1

(4)

v

T

mg

u A

B

Conservation of Energy 1 m  u 2  v 2   mgr 2 v 2  u 2  2 gr R  T

T  mg 

mv 2 r

m 2 u  2 gr   mg r mu 2   3mg r mgl   3mg 2r mgl  3mg 2r 1  r 6 5l ABMIN  6

T 0

r

M1 A1

M1 A1 M1 A1 M1

M1

A1

(9) [13]

8

9

10

11